極值的應用
Author: 任揚龢

$A(t)=\frac{1}{2}(2t+4)(4-t)$
           $=(t+2)(4-t²)$
           $=-t-2t²+4t+8$

$ A'(t)=-3t²-4t+4 $
            $=-(3t²+4t-4)$
            $=-(3t-2)(t+2)$

$ ⇒t=\frac{2}{3} $

$ A(\frac{2}{3})=\frac{8}{3}×\frac{32}{9} $
            $=\frac{256}{27}$

u = Vector((x(Corner[1]), 0), (x(Corner[3]), 0))
v = Vector((0, y(Corner[1])), (0, y(Corner[3])))
t = 0.62 (slider from 0 to 2, step 0.01)
A, B = Intersect(c, xAxis)
C: (t, 4 - t^(2))
D: (-t, 4 - t^(2))
poly1, a, b, c_1, d = Polygon(A, B, C, D)
文字poly1: "Area " + (Name[A]) + (Name[B]) + (Name[C]) + (Name[D]) + " = " + poly1